xorl %eax, %eax in x86_64 assembly code produced by gcc

问题

I'm a total noob at assembly, just poking around a bit to see what's going on. Anyway, I wrote a very simple function:

void multA(double *x,long size)
{
  long i;
  for(i=0; i<size; ++i){
    x[i] = 2.4*x[i];
  }
}

I compiled it with:

gcc -S -m64 -O2 fun.c

And I get this:

    .file   "fun.c"
    .text
    .p2align 4,,15
    .globl  multA
    .type   multA, @function
multA:
.LFB34:
    .cfi_startproc
    testq   %rsi, %rsi
    jle .L1
    movsd   .LC0(%rip), %xmm1
    xorl    %eax, %eax
    .p2align 4,,10
    .p2align 3
.L3:
    movsd   (%rdi,%rax,8), %xmm0
    mulsd   %xmm1, %xmm0
    movsd   %xmm0, (%rdi,%rax,8)
    addq    $1, %rax
    cmpq    %rsi, %rax
    jne .L3
.L1:
    rep
    ret
    .cfi_endproc
.LFE34:
    .size   multA, .-multA
    .section    .rodata.cst8,"aM",@progbits,8
    .align 8
.LC0:
    .long   858993459
    .long   1073951539
    .ident  "GCC: (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3"
    .section    .note.GNU-stack,"",@progbits

The assembly output makes sense to me (mostly) except for the line xorl %eax, %eax. From googling, I gather that the purpose of this is simply to set %eax to zero, which in this case corresponds to my iterator long i;.

However, unless I am mistaken, %eax is a 32-bit register. So it seems to me that this should actually be xorq %rax, %rax, particularly since this is holding a 64-bit long int. Moreover, further down in the code, it actually uses the 64-bit register %rax to do the iterating, which never gets initialized outside of xorl %eax %eax, which would seem to only zero out the lower 32 bits of the register.

Am I missing something?

Also, out of curiosity, why are there two .long constants there at the bottom? The first one, 858993459 is equal to the double floating-point representation of 2.4 but I can't figure out what the second number is or why it is there.

回答1:

I gather that the purpose of this is simply to set %eax to zero

Yes.

which in this case corresponds to my iterator long i;.

No. Your i is uninitialized in the declaration. Strictly speaking, that operation corresponds to the i = 0 expression in the for loop.

However, unless I am mistaken, %eax is a 32-bit register. So it seems to me that this should actually be xorq %rax, %rax, particularly since this is holding a 64-bit long int.

But clearing the lower double word of the register clears the entire register. This is not intuitive, but it's implicit.

回答2:

Just to answer the second part: .long means 32 bit, and the two integral constants side-by-side form the IEEE-754 representation of the double 2.4:

Dec: 1073951539  858993459
Hex: 0x40033333 0x33333333

     400 3333333333333
     S+E Mantissa

The exponent is offset by 1023, so the actual exponent is 0x400 − 1023 = 1. The leading "one" in the mantissa is implied, so it's 2¹ × 0b1.001100110011... (You recognize this periodic expansion as 3/15, i.e. 0.2. Sure enough, 2 × 1.2 = 2.4.)

来源：https://stackoverflow.com/questions/19082108/xorl-eax-eax-in-x86-64-assembly-code-produced-by-gcc

标签

gcc

assembly

x86-64